Physics - Electrostatics Question with Solution | TestHub

The capacitor is charged by a battery$25 \mathrm{V}$. Let the magnitude of surface charge density on each plate be$\sigma$. Before inserting the dielectric slab, field strength between the plates,

$E=\frac{\sigma }{{\epsilon }_0}=\frac{V}{d}$

or$E=\frac{\sigma }{{\epsilon }_0}=\frac{25}{5\times 10^{-3}}=5000 \mathrm{N}/\mathrm{C}$

The capacitor is disconnected from the battery but the charge on it will not change so$\sigma$has the same value. When a dielectric slab of thickness$3mm$is placed between the plates, the air thickness between the plates will be$5-3=2 \mathrm{mm}$. Electric field strength in air will have the same value$\left(5000 \mathrm{N}/\mathrm{C}\right)$but inside the dielectric, it will be$\frac{5000}{K}=\frac{5000}{10}=500 \mathrm{N}/\mathrm{C}$

So potential difference

$V=E_{air}{d}_{air}+E_{med}{d}_{med}$

$V=5000\times \left(2\times 10^{-3}\right)+500\times \left(3\times 10^{-3}\right)$

$V=11.5 \mathrm{V}$